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As in another question, let

$$T(N) = \begin{cases}1 & \text{if } N = 1\\ T(\phi(N)) + \lg(\phi(N))^3 & \text{otherwise} \end{cases}$$

where $\phi(N)$ is Euler's totient function.

Tasse kindly reasoned that we can write

$$T(N) = T(\phi(N))+\log(\phi(N))^3 = T(\phi(\phi(N)))+\log(\phi(\phi(N)))^3 + \log(\phi(N))^3$$

I can't at all see why this is true. How is this equation derived? I do understand his proof that $\phi(\phi(N)) < N/2$.

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Here's the relevant part of the proof, quoting verbatim:

First, show that $\phi(\phi(n)) < n/2$. This can be done as such:

Let $n = \prod_{i=1}^rp_i^{k_i}$ be the prime factorisation of $n$ ($p_i$ prime, $k_i>0$)

  • Suppose $n$ is even. Then $\phi(n) = n\prod_{i=1}^r(1-\frac{1}{p_i}) \leq n(1-\frac{1}{2}) \leq n/2.$ Thus $\phi(\phi(n)) < n/2$.
  • Suppose $n$ is odd and $n > 1$. Then $\phi(n) = \prod_{i=1}^r (p_i-1)p_i^{k_i-1}$ is even and smaller than $n$. By the previous result $\phi(\phi(n)) < n/2$.

So we get the desired result.

This lemma is slightly mis-stated, because when $n=2$, $\phi(\phi(n)) = n/2$. But it doesn't invalidate the main result. We will state the lemma as:

Lemma For all $n \ge 2$, $\phi(\phi(n)) \le \frac{n}{2}$.

Tasse's proof sketch is a proof by induction in disguise, but this may not have been obvious to you. I'm going to simplify the proof sketch a little bit and lay it out more like a proper proof by induction.

Proof By induction on $n$.

Base case: If $n=2$, see above.

Inductive step: Suppose that for all $2 \le m < n$, $\phi(\phi(m)) \le \frac{m}{2}$. We want to prove that $\phi(\phi(n)) \le \frac{n}{2}$.

Case 1: Suppose $n$ is even, that is, $n = 2q$ for some $q$. In this case, the multiplicative property of the totient function says that $\phi(2q) = \phi(2) \phi(q) = \phi(q)$.

That is, $\phi(n) = \phi(\frac{n}{2})$. However, another useful property of the totient function is that for all $m \ge 2$, $\phi(m) \le m-1$. Therefore $\phi(\phi(n)) \le \frac{n}{2} - 1$.

This proves what we wanted for this case, but we have also incidentally proven another useful thing: If $n$ is even, then $\phi(n) \le \frac{n}{2}$. This will be important in the next step.

Case 2: Suppose $n$ is odd. That is, $n = p^k q$ for some odd prime $p$, and some positive integer $k$, and some $q$ which doesn't have $p$ as a divisor. Then:

$\phi(n) = \phi(p^k) \phi(q) = \left(p^k - p^{k-1}\right) \phi(q)$.

Now there are two things you need to notice about the right-hand side.

First off, it's less than $n$. Why? Because clearly $p^k - p^{k-1} < p^k$, and it's a standard property of the totient function that $\phi(q) \le q$. (Note that this is also true if $q = 1$.)

Secondly, it's even. Why? Because if $p$ is odd, $p^k - p^{k-1}$ is even.

So by that incidental result we discovered in case 1:

$\phi(\phi(n)) = \phi(\left(p^k - p^{k-1}\right) \phi(q)) \le \frac{\left(p^k - p^{k-1}\right) \phi(q)}{2} < \frac{n}{2}$.

QED

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  • $\begingroup$ It looks like you've clarified quite a bit the result that $\phi(\phi(N)) \le N/2$, and I will dutifully review it, but the question is actually about the equation $$T(N) = T(\phi(N))+\log(\phi(N))^3 = T(\phi(\phi(N)))+\log(\phi(\phi(N)))^3 + \log(\phi(N))^3,$$ which is what I don't have any idea why it's true. If you could explain it with the level of clarity you seem to explain the lemma, that'd be really great. Thanks! $\endgroup$ – R. Chopin Oct 1 '19 at 13:15
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    $\begingroup$ Oh, sorry, I clarified the wrong part! I'll take a look tomorrow. $\endgroup$ – Pseudonym Oct 1 '19 at 13:36
  • $\begingroup$ Pseudonym, $\phi(2q) = \phi(2)\phi(q)$ provided $\gcd(2,q) = 1$, but all we have in your case 1 is that $2q$ is even; we don't know $q$ is odd, so you cannot use the multiplicative property of $\phi$. What do you say? $\endgroup$ – R. Chopin Oct 2 '19 at 14:07

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